Question

For this graphical display of Quiz Scores, which estimates of the mean and median are most plausible? 3 2 6 10 14 18 22 uiz Seores median = 13.0 and mean = 12.0 median 14.0 and mean 15.0 median16.0 and mean 14.3 median = 16.5 and mean = 16.2

Answer 1

Given is a graphical display of Quiz Scores, now we need to find which estimates of the mean and median are most plausible.

Now clearly the the graphical display is a histogram with classes of length 2 in the x axis and the corresponding frequencies in the y axis. Now let us re-make the frequency distribution which corresponds to the histogram.

No.	Class Boundary		Frequency
	Lower	Upper
1	4	6	1
2	6	8	1
3	8	10	0
4	10	12	1
5	12	14	2
6	14	16	3
7	16	18	5
8	18	20	4
Total			17

Now let us calculate the median of this frequency distribution. For that we prepare a cumulative frequency column of less than type.

No.	Class Boundary		Frequency	Cumulative Frequency
	Lower	Upper
1	4	6	1	1
2	6	8	1	2
3	8	10	0	2
4	10	12	1	3
5	12	14	2	5
6	14	16	3	8
7	16	18	5	13
8	18	20	4	17
Total			17

Now median is given by,

$Median=x_{l}+\frac{n/2-n_{l}}{f_{0}}*c$

$x_{l}=\emph{Lower Class Boundary of the class containing the median}$

$n=\emph{Total frequency}$

$n_{l}=\emph{Cumulative frequency of the previous class containing the median}$

$f_{0}=\emph{Frequency of the class containing the median}$

$c=\emph{Class Width}$

Now here the median class is, n/2=17/2=8.5 i.e the 7th class is the median class.

Hence,

$x_{l}=16,n=17,n_{l}=8,f_{0}=5,c=2$

$\Rightarrow Median=16+\frac{17/2-8}{5}*2$

$\Rightarrow Median=16+0.2$

$\Rightarrow Median=16.2$

Now mean is given by,

$Mean=\frac{1}{n}\sum_{i=1}^{k}x_{i}f_{i}$

where,

$n=\sum_{i=1}^{k}f_{i}$

$x_{i}=\emph{Class Mark of the ith class}$

$f_{i}=\emph{Frequency of the ith class}$

$k=\emph{Number of classes}$

So we make a class mark column our frequency distribution. Now class mark is the average of the upper and class boundary.

No.	Class Boundary		Class Mark(xi)	Frequency(fi)	xi*fi
	Lower	Upper
1	4	6	5	1	5
2	6	8	7	1	7
3	8	10	9	0	0
4	10	12	11	1	11
5	12	14	13	2	26
6	14	16	15	3	45
7	16	18	17	5	85
8	18	20	19	4	76
Total				17	255

$Mean=\frac{1}{n}\sum_{i=1}^{k}x_{i}f_{i}=\frac{255}{17}$.

$\Rightarrow Mean=15$

Hence Median=16.2 and Mean=15.

Hence the most plausible option is third option Median=16 and Mean=14.3. Since they are close to the exact mean and median.